3 LEAST UPPER BOUND AND GREATEST LOWER BOUND 25

(2.3.6) Let (AA)AeL be a family of nonempty majorized subsets of R; let
A = (J AA, and let B be the set of elements sup AA. In order that A be major-

XeL

ized, a necessary and sufficient condition is that B be majorized, and then
sup A = sup B.

It follows at once from the definition that any majorant of A is a majorant
of B, and vice versa, hence the result.

Let/be a mapping of a set A into the set R of real numbers;/is said
to be majorized (resp. minorized, bounded) in A if the subset /(A) of
R is majorized (resp. minorized, bounded); we write sup/(A) = sup/(x),

JceA

inf/(A) = inf f(x) when these numbers are defined (supremum and infimum

xeA

off in A). If/is majorized, then —/is minorized, and
inf (-/(*)) =-sup/(x).

xeA xeA

(2.3.7) Let f be a mapping ofAl x A2 into R; iff is majorized,

sup /(*!, x2) = sup ( sup /(*!, x2))

Oil *2>e AI x AS xi eAi *26 Aa

For we can write /(Ai x A2) as the union of sets/^} x A2), jq ranging
through Al9 and apply (2.3.6).

(2.3.8) Letf, g be two mappings of A into R such thatf(x) < g(x)for every
xeA; then ifg is majorized, so isf, and sup/(x) < sup g(x).

JceA

This follows immediately from the definitions.

(2.3.9) Let f and g be two mappings of A into R; if f and g are both major-
ized, so is f+ g (i.e. the mapping x-*f(x) +#(;*:)), and

sup (/(*) + g(x)) < sup f(x) 4- sup g (x).

xeA xeA xeA

If in addition g is minorized then

sup/(x) 4- inf g(x) < sup (f(x) + g(x))

xeA xeA xeA
are satisfied, we cannot have sup X = /? < y, for there would be an n such